Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}4x+8y &= 8 \\ -3x-y &= -6\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = y-6$ Divide both sides by $-3$ to isolate $x$ $x = {-\dfrac{1}{3}y + 2}$ Substitute this expression for $x$ in the first equation. $4({-\dfrac{1}{3}y + 2}) + 8y = 8$ $-\dfrac{4}{3}y + 8 + 8y = 8$ Simplify by combining terms, then solve for $y$ $\dfrac{20}{3}y + 8 = 8$ $\dfrac{20}{3}y = 0$ $y = 0$ Substitute $0$ for $y$ in the top equation. $4x+8( 0) = 8$ $4x = 8$ $4x = 8$ $x = 2$ The solution is $\enspace x = 2, \enspace y = 0$.